384
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
(15.1")
sinnx
(~) COS n - 1 x. sinx( n) n-3 . 3
- 3 cos
x. sm X + ... ;
to obtain these one expands (cos x ± i sin x)n by the binomial formula and
adds or subtracts the two formulae so obtained. It seems that they are not
there literally in de Moivre, and it was Euler who, by this method, wrote
them in his Introductio. And when one is called Euler, one deduces, why not,
the expansions in power series of cos x and sin x, whence, starting from these,
another proof of the relations (1 ') and (1"). The method, we see, is again of
genius, - and, again, finesses the essential point.
Here it is: in (1'), for x given, you replace x by x/n with n infinitely
large and so x/n infinitely small; the left hand side becomes cos x; on the
right hand side you replace cos(x/n) and its powers by 1 since cosO = 1, and
sin(x/n) by x/n since sin u "" u for u infinitely small; sink(x/n) then becomes
xk Ink and third term of (1 ') for example becomes
n(n -I)(n - 2)(n - 3)x4/n4.4! = 1(1 -I/n)(1- 2/n)(1 - 3/n)x 4 /4!.
But since n is infinitely large, I/n, 2/n, etc. are zero, and the preceding
expression reduces to x 4 /4!; and so one (re)finds Newton's series cosx =
1- x 2 /2! +x 4 /4! - ...
Once again we have to deal with a passage to the limit over n in a sum
whose number of terms is certainly finite, but increases indefinitely with n.
This suggests using the dominated convergence theorem of n° 12, interchanging the letters n and p since here we sum over p while passing to the limit
over n.
In the general term
un(p) = n(n - 1) ... (n - 2p) cosn- 2 p-1(x/n). sin 2p + 1 (x/n)/(2p + I)!,
let us put a factor n in each of 2p + 1 terms of the form n - k and gather to
the sine the product of these factors n; we find
(15.2) (1 - I/n) ... (1 - 2p/n) cosn- 2 p-1(x/n)[n. sin(x/n)]2 p +l /(2p + I)!.
By Theorem 11, we need only find a convergent series Ev(p) with positive
terms such that v(p) majorises the expression above for any n, and then show
that this tends to x 2p +1 /(2p + I)! as n increases indefinitely.
It is easy to find v(p). Clearly
IUn (p) I ::; [no sin(x/n)]2p+l /(2p + I)!.
As we have seen above, n. sin(x/n) tends to x as n increases, so is, for x
given, majorised in modulus by a constant M(x) whose value is not imporant
because we then obtain the inequality
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
(15.1")
sinnx
(~) COS n - 1 x. sinx( n) n-3 . 3
- 3 cos
x. sm X + ... ;
to obtain these one expands (cos x ± i sin x)n by the binomial formula and
adds or subtracts the two formulae so obtained. It seems that they are not
there literally in de Moivre, and it was Euler who, by this method, wrote
them in his Introductio. And when one is called Euler, one deduces, why not,
the expansions in power series of cos x and sin x, whence, starting from these,
another proof of the relations (1 ') and (1"). The method, we see, is again of
genius, - and, again, finesses the essential point.
Here it is: in (1'), for x given, you replace x by x/n with n infinitely
large and so x/n infinitely small; the left hand side becomes cos x; on the
right hand side you replace cos(x/n) and its powers by 1 since cosO = 1, and
sin(x/n) by x/n since sin u "" u for u infinitely small; sink(x/n) then becomes
xk Ink and third term of (1 ') for example becomes
n(n -I)(n - 2)(n - 3)x4/n4.4! = 1(1 -I/n)(1- 2/n)(1 - 3/n)x 4 /4!.
But since n is infinitely large, I/n, 2/n, etc. are zero, and the preceding
expression reduces to x 4 /4!; and so one (re)finds Newton's series cosx =
1- x 2 /2! +x 4 /4! - ...
Once again we have to deal with a passage to the limit over n in a sum
whose number of terms is certainly finite, but increases indefinitely with n.
This suggests using the dominated convergence theorem of n° 12, interchanging the letters n and p since here we sum over p while passing to the limit
over n.
In the general term
un(p) = n(n - 1) ... (n - 2p) cosn- 2 p-1(x/n). sin 2p + 1 (x/n)/(2p + I)!,
let us put a factor n in each of 2p + 1 terms of the form n - k and gather to
the sine the product of these factors n; we find
(15.2) (1 - I/n) ... (1 - 2p/n) cosn- 2 p-1(x/n)[n. sin(x/n)]2 p +l /(2p + I)!.
By Theorem 11, we need only find a convergent series Ev(p) with positive
terms such that v(p) majorises the expression above for any n, and then show
that this tends to x 2p +1 /(2p + I)! as n increases indefinitely.
It is easy to find v(p). Clearly
IUn (p) I ::; [no sin(x/n)]2p+l /(2p + I)!.
As we have seen above, n. sin(x/n) tends to x as n increases, so is, for x
given, majorised in modulus by a constant M(x) whose value is not imporant
because we then obtain the inequality